The system of equations graphed below has how many solutions? y= 2x+2, y=2x
2 answers:
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Answer:
The 95% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).
Step-by-step explanation:
Our sample size is 96.
The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So
Then, we need to subtract one by the confidence level and divide by 2. So:
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 95 and 0.025 in the t-distribution table, we have .
Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So
Now, we multiply T and s
For the lower end of the interval, we subtract the mean by M. So
For the upper end of the interval, we add the mean to M. So
The 95% confidence interval estimate for the mean highway mileage for SUVs is (18.29mpg, 20.91mpg).
Answer:
a) 0.259
b) 0.297
c) 0.497
Step-by-step explanation:
To solve this problem, it is important to know the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.6293.
X = 10.8
has a pvalue of 0.3707.
0.6293 - 0.3707 = 0.2586
0.259 probability, rounded to three decimal places.
b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?
Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So
X = 11
has a pvalue of 0.5.
X = 10.5
has a pvalue of 0.2033.
0.5 - 0.2033 = 0.2967
0.297, rounded to three decimal places.
c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?
Here we have that
This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.
X = 11.2
By the Central Limit Theorem
has a pvalue of 0.7486.
X = 10.8
has a pvalue of 0.2514.
0.7486 - 0.2514 = 0.4972
0.497, rounded to three decimal places.